How to show $1 + t^p geq (1+t)^p$ for $p 0$?
$begingroup$
How can we show $$1 + t^p geq (1+t)^p$$
where $p < 1$ and $t>0$?
This inequality is useful to show $sqrt{a+b} leq sqrt{a}+ sqrt{b}$ for $a,bgeq 0$.
My try:
One can define $f(t)=1+t^p-(1+t)^p$ so $f'(t)=pt^{p-1}-p(1+t)^{p-1}$.
Therefore,
$$f'(t)=p(t^{p-1}-(1+t)^{p-1})=p(frac{1}{t^{1-p}}-frac{1}{(1+t)^{1-p}})$$
$$
t < 1 +t rightarrow frac{1}{t} > frac{1}{1 +t}
$$
Now I am wondering what is the general rule for raising both sides of an equality to a positive number?
Is it true $|a|<|b| rightarrow |a|^p<|b|^p ,,, forall p>0$?
Can you help me to figure out the rule when both sides are not positive?
functions inequality
asked Jan 16 at 16:28
SaeedSaeed
1,005310
$endgroup$
add a comment |
$begingroup$
How can we show $$1 + t^p geq (1+t)^p$$
where $p < 1$ and $t>0$?
This inequality is useful to show $sqrt{a+b} leq sqrt{a}+ sqrt{b}$ for $a,bgeq 0$.
My try:
One can define $f(t)=1+t^p-(1+t)^p$ so $f'(t)=pt^{p-1}-p(1+t)^{p-1}$.
Therefore,
$$f'(t)=p(t^{p-1}-(1+t)^{p-1})=p(frac{1}{t^{1-p}}-frac{1}{(1+t)^{1-p}})$$
$$
t < 1 +t rightarrow frac{1}{t} > frac{1}{1 +t}
$$
Now I am wondering what is the general rule for raising both sides of an equality to a positive number?
Is it true $|a|<|b| rightarrow |a|^p<|b|^p ,,, forall p>0$?
Can you help me to figure out the rule when both sides are not positive?
functions inequality
asked Jan 16 at 16:28
SaeedSaeed
1,005310
$endgroup$
2
$begingroup$
For your original inequality, you can take the square of both sides. For your question, I suggest that you use the slope inequalities and recall that $x longmapsto x^p$ is concave.
$endgroup$
– Mindlack
Jan 16 at 16:32
$begingroup$
@Mindlack: When $p=2$ it is not concave, it is convex?
$endgroup$
– Saeed
Jan 16 at 16:37
1
$begingroup$
You specified that $p < 1$!
$endgroup$
– Mindlack
Jan 16 at 16:43
$begingroup$
@Mindlack: It is concave but it is still increasing? What is the point of concavity? Now, is it true $|a|<|b| rightarrow |a|^p<|b|^p ,,, forall p>0$? because if we take the the derivative of $f(x)=x^p$ is $f'(x)=px^{p-1}$ which has all positive element and is increasing.
$endgroup$
– Saeed
Jan 16 at 17:01
$begingroup$
Yes, the function is concave increasing. Note that its derivative decreases because $p <1$.
$endgroup$
– Mindlack
Jan 16 at 21:59
add a comment |
$begingroup$
How can we show $$1 + t^p geq (1+t)^p$$
where $p < 1$ and $t>0$?
This inequality is useful to show $sqrt{a+b} leq sqrt{a}+ sqrt{b}$ for $a,bgeq 0$.
My try:
One can define $f(t)=1+t^p-(1+t)^p$ so $f'(t)=pt^{p-1}-p(1+t)^{p-1}$.
Therefore,
$$f'(t)=p(t^{p-1}-(1+t)^{p-1})=p(frac{1}{t^{1-p}}-frac{1}{(1+t)^{1-p}})$$
$$
t < 1 +t rightarrow frac{1}{t} > frac{1}{1 +t}
$$
Now I am wondering what is the general rule for raising both sides of an equality to a positive number?
Is it true $|a|<|b| rightarrow |a|^p<|b|^p ,,, forall p>0$?
Can you help me to figure out the rule when both sides are not positive?
functions inequality
asked Jan 16 at 16:28
SaeedSaeed
1,005310
$endgroup$
How can we show $$1 + t^p geq (1+t)^p$$
where $p < 1$ and $t>0$?
This inequality is useful to show $sqrt{a+b} leq sqrt{a}+ sqrt{b}$ for $a,bgeq 0$.
My try:
One can define $f(t)=1+t^p-(1+t)^p$ so $f'(t)=pt^{p-1}-p(1+t)^{p-1}$.
Therefore,
$$f'(t)=p(t^{p-1}-(1+t)^{p-1})=p(frac{1}{t^{1-p}}-frac{1}{(1+t)^{1-p}})$$
$$
t < 1 +t rightarrow frac{1}{t} > frac{1}{1 +t}
$$
Now I am wondering what is the general rule for raising both sides of an equality to a positive number?
Is it true $|a|<|b| rightarrow |a|^p<|b|^p ,,, forall p>0$?
Can you help me to figure out the rule when both sides are not positive?
functions inequality
functions inequality
asked Jan 16 at 16:28
SaeedSaeed
1,005310
asked Jan 16 at 16:28
SaeedSaeed
1,005310
edited Jan 16 at 16:35
Saeed
asked Jan 16 at 16:28
SaeedSaeed
1,005310
asked Jan 16 at 16:28
SaeedSaeed
1,005310
asked Jan 16 at 16:28
SaeedSaeed
1,005310
1,005310
2
$begingroup$
For your original inequality, you can take the square of both sides. For your question, I suggest that you use the slope inequalities and recall that $x longmapsto x^p$ is concave.
$endgroup$
– Mindlack
Jan 16 at 16:32
$begingroup$
@Mindlack: When $p=2$ it is not concave, it is convex?
$endgroup$
– Saeed
Jan 16 at 16:37
1
$begingroup$
You specified that $p < 1$!
$endgroup$
– Mindlack
Jan 16 at 16:43
$begingroup$
@Mindlack: It is concave but it is still increasing? What is the point of concavity? Now, is it true $|a|<|b| rightarrow |a|^p<|b|^p ,,, forall p>0$? because if we take the the derivative of $f(x)=x^p$ is $f'(x)=px^{p-1}$ which has all positive element and is increasing.
$endgroup$
– Saeed
Jan 16 at 17:01
$begingroup$
Yes, the function is concave increasing. Note that its derivative decreases because $p <1$.
$endgroup$
– Mindlack
Jan 16 at 21:59
add a comment |
2
$begingroup$
For your original inequality, you can take the square of both sides. For your question, I suggest that you use the slope inequalities and recall that $x longmapsto x^p$ is concave.
$endgroup$
– Mindlack
Jan 16 at 16:32
$begingroup$
@Mindlack: When $p=2$ it is not concave, it is convex?
$endgroup$
– Saeed
Jan 16 at 16:37
1
$begingroup$
You specified that $p < 1$!
$endgroup$
– Mindlack
Jan 16 at 16:43
$begingroup$
@Mindlack: It is concave but it is still increasing? What is the point of concavity? Now, is it true $|a|<|b| rightarrow |a|^p<|b|^p ,,, forall p>0$? because if we take the the derivative of $f(x)=x^p$ is $f'(x)=px^{p-1}$ which has all positive element and is increasing.
$endgroup$
– Saeed
Jan 16 at 17:01
$begingroup$
Yes, the function is concave increasing. Note that its derivative decreases because $p <1$.
$endgroup$
– Mindlack
Jan 16 at 21:59
2
2
$begingroup$
For your original inequality, you can take the square of both sides. For your question, I suggest that you use the slope inequalities and recall that $x longmapsto x^p$ is concave.
$endgroup$
– Mindlack
Jan 16 at 16:32
$begingroup$
For your original inequality, you can take the square of both sides. For your question, I suggest that you use the slope inequalities and recall that $x longmapsto x^p$ is concave.
$endgroup$
– Mindlack
Jan 16 at 16:32
$begingroup$
@Mindlack: When $p=2$ it is not concave, it is convex?
$endgroup$
– Saeed
Jan 16 at 16:37
$begingroup$
@Mindlack: When $p=2$ it is not concave, it is convex?
$endgroup$
– Saeed
Jan 16 at 16:37
1
1
$begingroup$
You specified that $p < 1$!
$endgroup$
– Mindlack
Jan 16 at 16:43
$begingroup$
You specified that $p < 1$!
$endgroup$
– Mindlack
Jan 16 at 16:43
$begingroup$
@Mindlack: It is concave but it is still increasing? What is the point of concavity? Now, is it true $|a|<|b| rightarrow |a|^p<|b|^p ,,, forall p>0$? because if we take the the derivative of $f(x)=x^p$ is $f'(x)=px^{p-1}$ which has all positive element and is increasing.
$endgroup$
– Saeed
Jan 16 at 17:01
$begingroup$
@Mindlack: It is concave but it is still increasing? What is the point of concavity? Now, is it true $|a|<|b| rightarrow |a|^p<|b|^p ,,, forall p>0$? because if we take the the derivative of $f(x)=x^p$ is $f'(x)=px^{p-1}$ which has all positive element and is increasing.
$endgroup$
– Saeed
Jan 16 at 17:01
$begingroup$
Yes, the function is concave increasing. Note that its derivative decreases because $p <1$.
$endgroup$
– Mindlack
Jan 16 at 21:59
$begingroup$
Yes, the function is concave increasing. Note that its derivative decreases because $p <1$.
$endgroup$
– Mindlack
Jan 16 at 21:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $t=0$, equality holds. Now for $t>0$, look at the derivatives w.r.t. $t$ on both sides:
$$
(1+t^p)' = p t^{p-1} \
((1+t)^p)' = p (1+t)^{p-1}
$$
As $0< p<1$, we have for all $t$ that
$$ t < 1+ t \
log t < log (1+t)\
(p-1) log t > (p-1) log (1+t) \
t^{p-1} > (1+t)^{p-1} \
p t^{p-1} > p (1+t)^{p-1} \
(1+t^p)' >
((1+t)^p)' $$
so, starting from equality at $t=0$, $1+t^p$ grows faster with $t$ than $(1+t)^p$ (for all $0<p<1$) which proves the claim.
answered Jan 16 at 20:57
AndreasAndreas
8,1131037
$endgroup$
$begingroup$
What are you doing to get $ t^{p-1} > (1+t)^{p-1} $ from $(p-1) log t > (p-1) log (1+t)$?
$endgroup$
– Saeed
Jan 16 at 22:07
$begingroup$
Exponentiate both sides, e.g. $exp((p-1) log t ) = t^{p-1}$. Since the $exp$-function is strictly monotonously rising everywhere, exponentiating preserves the inequality.
$endgroup$
– Andreas
Jan 17 at 8:12
add a comment |
Your Answer
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$begingroup$
For $t=0$, equality holds. Now for $t>0$, look at the derivatives w.r.t. $t$ on both sides:
$$
(1+t^p)' = p t^{p-1} \
((1+t)^p)' = p (1+t)^{p-1}
$$
As $0< p<1$, we have for all $t$ that
$$ t < 1+ t \
log t < log (1+t)\
(p-1) log t > (p-1) log (1+t) \
t^{p-1} > (1+t)^{p-1} \
p t^{p-1} > p (1+t)^{p-1} \
(1+t^p)' >
((1+t)^p)' $$
so, starting from equality at $t=0$, $1+t^p$ grows faster with $t$ than $(1+t)^p$ (for all $0<p<1$) which proves the claim.
answered Jan 16 at 20:57
AndreasAndreas
8,1131037
$endgroup$
$begingroup$
What are you doing to get $ t^{p-1} > (1+t)^{p-1} $ from $(p-1) log t > (p-1) log (1+t)$?
$endgroup$
– Saeed
Jan 16 at 22:07
$begingroup$
Exponentiate both sides, e.g. $exp((p-1) log t ) = t^{p-1}$. Since the $exp$-function is strictly monotonously rising everywhere, exponentiating preserves the inequality.
$endgroup$
– Andreas
Jan 17 at 8:12
add a comment |
$begingroup$
For $t=0$, equality holds. Now for $t>0$, look at the derivatives w.r.t. $t$ on both sides:
$$
(1+t^p)' = p t^{p-1} \
((1+t)^p)' = p (1+t)^{p-1}
$$
As $0< p<1$, we have for all $t$ that
$$ t < 1+ t \
log t < log (1+t)\
(p-1) log t > (p-1) log (1+t) \
t^{p-1} > (1+t)^{p-1} \
p t^{p-1} > p (1+t)^{p-1} \
(1+t^p)' >
((1+t)^p)' $$
so, starting from equality at $t=0$, $1+t^p$ grows faster with $t$ than $(1+t)^p$ (for all $0<p<1$) which proves the claim.
answered Jan 16 at 20:57
AndreasAndreas
8,1131037
$endgroup$
$begingroup$
What are you doing to get $ t^{p-1} > (1+t)^{p-1} $ from $(p-1) log t > (p-1) log (1+t)$?
$endgroup$
– Saeed
Jan 16 at 22:07
$begingroup$
Exponentiate both sides, e.g. $exp((p-1) log t ) = t^{p-1}$. Since the $exp$-function is strictly monotonously rising everywhere, exponentiating preserves the inequality.
$endgroup$
– Andreas
Jan 17 at 8:12
add a comment |
$begingroup$
For $t=0$, equality holds. Now for $t>0$, look at the derivatives w.r.t. $t$ on both sides:
$$
(1+t^p)' = p t^{p-1} \
((1+t)^p)' = p (1+t)^{p-1}
$$
As $0< p<1$, we have for all $t$ that
$$ t < 1+ t \
log t < log (1+t)\
(p-1) log t > (p-1) log (1+t) \
t^{p-1} > (1+t)^{p-1} \
p t^{p-1} > p (1+t)^{p-1} \
(1+t^p)' >
((1+t)^p)' $$
so, starting from equality at $t=0$, $1+t^p$ grows faster with $t$ than $(1+t)^p$ (for all $0<p<1$) which proves the claim.
answered Jan 16 at 20:57
AndreasAndreas
8,1131037
$endgroup$
For $t=0$, equality holds. Now for $t>0$, look at the derivatives w.r.t. $t$ on both sides:
$$
(1+t^p)' = p t^{p-1} \
((1+t)^p)' = p (1+t)^{p-1}
$$
As $0< p<1$, we have for all $t$ that
$$ t < 1+ t \
log t < log (1+t)\
(p-1) log t > (p-1) log (1+t) \
t^{p-1} > (1+t)^{p-1} \
p t^{p-1} > p (1+t)^{p-1} \
(1+t^p)' >
((1+t)^p)' $$
so, starting from equality at $t=0$, $1+t^p$ grows faster with $t$ than $(1+t)^p$ (for all $0<p<1$) which proves the claim.
answered Jan 16 at 20:57
AndreasAndreas
8,1131037
answered Jan 16 at 20:57
AndreasAndreas
8,1131037
answered Jan 16 at 20:57
AndreasAndreas
8,1131037
answered Jan 16 at 20:57
AndreasAndreas
8,1131037
8,1131037
$begingroup$
What are you doing to get $ t^{p-1} > (1+t)^{p-1} $ from $(p-1) log t > (p-1) log (1+t)$?
$endgroup$
– Saeed
Jan 16 at 22:07
$begingroup$
Exponentiate both sides, e.g. $exp((p-1) log t ) = t^{p-1}$. Since the $exp$-function is strictly monotonously rising everywhere, exponentiating preserves the inequality.
$endgroup$
– Andreas
Jan 17 at 8:12
add a comment |
$begingroup$
What are you doing to get $ t^{p-1} > (1+t)^{p-1} $ from $(p-1) log t > (p-1) log (1+t)$?
$endgroup$
– Saeed
Jan 16 at 22:07
$begingroup$
Exponentiate both sides, e.g. $exp((p-1) log t ) = t^{p-1}$. Since the $exp$-function is strictly monotonously rising everywhere, exponentiating preserves the inequality.
$endgroup$
– Andreas
Jan 17 at 8:12
$begingroup$
What are you doing to get $ t^{p-1} > (1+t)^{p-1} $ from $(p-1) log t > (p-1) log (1+t)$?
$endgroup$
– Saeed
Jan 16 at 22:07
$begingroup$
What are you doing to get $ t^{p-1} > (1+t)^{p-1} $ from $(p-1) log t > (p-1) log (1+t)$?
$endgroup$
– Saeed
Jan 16 at 22:07
$begingroup$
Exponentiate both sides, e.g. $exp((p-1) log t ) = t^{p-1}$. Since the $exp$-function is strictly monotonously rising everywhere, exponentiating preserves the inequality.
$endgroup$
– Andreas
Jan 17 at 8:12
$begingroup$
Exponentiate both sides, e.g. $exp((p-1) log t ) = t^{p-1}$. Since the $exp$-function is strictly monotonously rising everywhere, exponentiating preserves the inequality.
$endgroup$
– Andreas
Jan 17 at 8:12
add a comment |
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$begingroup$
For your original inequality, you can take the square of both sides. For your question, I suggest that you use the slope inequalities and recall that $x longmapsto x^p$ is concave.
$endgroup$
– Mindlack
Jan 16 at 16:32
$begingroup$
@Mindlack: When $p=2$ it is not concave, it is convex?
$endgroup$
– Saeed
Jan 16 at 16:37
1
$begingroup$
You specified that $p < 1$!
$endgroup$
– Mindlack
Jan 16 at 16:43
$begingroup$
@Mindlack: It is concave but it is still increasing? What is the point of concavity? Now, is it true $|a|<|b| rightarrow |a|^p<|b|^p ,,, forall p>0$? because if we take the the derivative of $f(x)=x^p$ is $f'(x)=px^{p-1}$ which has all positive element and is increasing.
$endgroup$
– Saeed
Jan 16 at 17:01
$begingroup$
Yes, the function is concave increasing. Note that its derivative decreases because $p <1$.
$endgroup$
– Mindlack
Jan 16 at 21:59